In a hostel, 60% of the students read Hindi newspaper,

Solution:

Let H denote the students who read Hindi newspaper and E denote the students who read English newspaper. It is given that,

$\mathrm{P}(\mathrm{H})=60 \%=\frac{6}{10}=\frac{3}{5}$

$\mathrm{P}(\mathrm{E})=40 \%=\frac{40}{100}=\frac{2}{5}$

$\mathrm{P}(\mathrm{H} \cap \mathrm{E})=20 \%=\frac{20}{100}=\frac{1}{5}$

(ii) Probability that a randomly chosen student reads English newspaper, if she reads Hindi news paper, is given by P (E|H

$P(E \mid H)=\frac{P(E \cap H)}{P(H)}$

$=\frac{\frac{1}{5}}{\frac{3}{5}}$

$=\frac{1}{3}$

(iii) Probability that a randomly chosen student reads Hindi newspaper, if she reads English newspaper, is given by P (H|E).

$P(H \mid E)=\frac{P(H \cap E)}{P(E)}$

$=\frac{1}{\frac{5}{2}}$

$=\frac{1}{5}$